50
Finite Control Volume Analysis

fealabreports
• Category

## Documents

• view

1.822

1

### Transcript of Finite CV Analysis

Finite Control Volume Analysis

Conservation of Mass – Continuity Equation

Reynolds transport theorem establishes relation between system rates of change and control-volume surface and volume integrals

With B = mass and b = 1, this becomes

Mass is conserved

Continuity equation

sys

sys0

DM DdV

Dt Dt

sys

cv cs

DBbdV b dA

Dt t

ˆV n

sys cv cs

DdV dV dA

Dt t

ˆV n

cv cs0dV dA

t

ˆV n

Conservation of Mass – Continuity Equation

Mass flowrate equals the product of density and volume flowrate

m Q AV

Fixed, Nondeforming Control Volume

Example 5.1 Seawater flows steadily through a simple conical-shaped nozzle at the end of a fire hose as illustrated in Figure. If the nozzle exit velocity must be at least 20 m/s, determine the minimum pumping capacity required in m3/s.

Example 5.2 Air flows steadily between two sections in a long, straight portion of 4-in. inside diameter pipe as indicated in Figure. The uniformly distributed temperature and pressure at each section are given. If the average air velocity (nonuniform velocity distribution) at section (2) is 1000 ft/s, calculate the average air velocity at section (1).

Example 5.3 Moist air (a mixture of dry air and water vapor) enters a dehumidifier at the rate of 22 slugs/hr. Liquid water drains out of the dehumidifier at a rate of 0.5 slugs/hr. Determine the mass flowrate of the dry air and the water vapor leaving the dehumidifier.

2 1 3Answer: 21.5 slugs/hrm m m

Example 5.4 Incompressible, laminar water flow develops in a straight pipe having radius R as indicated in Figure. At section (1), the velocity profile is uniform; the velocity is equal to a constant value U and is parallel to the pipe axis everywhere. At section (2), the velocity profile is axisymmetric and parabolic, with zero velocity at the pipe wall and a maximum value of umax at the centerline. How are U and umax related? How are the average velocity at

section (2), , and umax related?2V

2

uu U V

Example 5.5 A bathtub is being filled with water from a faucet. The rate of flow from the faucet is steady at 9 gal/min. The tub volume is approximated by a rectangular space as indicated in Figure. Estimate the time rate of change of the depth of water in the tub, h/t, in in./min at any instant.

water water2 2

Answer: 1.44 in./min10 ft 10 ftj

Q Qh

t A

Moving, Nondeforming Control Volume

Some problems are most easily solved using a moving control volume.

Continuity equation for a moving nondeforming control volume:

where relative velocity (fluid velocity seen by an observer moving with the control volume) is:

cv cs0dV dA

t

ˆW n

cv W V V

Example 5.6 An airplane moves forward at a speed of 971 km/hr. The frontal intake area of the jet engine is 0.80 m2 and the entering air density is 0.736 kg/m3. A stationary observerdetermines that relative to the earth, the jet engine exhaust gases move away from the engine with a speed of 1050 km/hr. The engine exhaust area is 0.558 m2, and the exhaustgas density is 0.515 kg/m3. Estimate the mass flowrate of fuel into the engine in kg/hr.

Example 5.7 Water enters a rotating lawn sprinkler through its base at the steady rate of 1000 ml/s. If the exit area of each of the two nozzles is 30 mm2, determine the average speed of the water leaving each nozzle, relative to the nozzle, if (a) the rotary sprinkler head is stationary, (b) the sprinkler head rotates at 600 rpm, and (c) the sprinkler head accelerates from 0 to 600 rpm.

22

QW

A

Deforming Control Volume

Some problems are most easily solved by using a deforming control volume.

Deforming control volume involves changing volume size and control surface movement.

Continuity equation for a deforming control volume:

For the deforming control volume

where Vcs is the velocity of the control surface as seen by a fixed observer.

cv cs0dV dA

t

ˆW n

cs V W V

Example 5.8 A syringe is used to inoculate a cow. The plunger has a face area of 500 mm2. If the liquid in the syringe is to be injected steadily at a rate of 300 cm3/min, at what speed should the plunger be advanced? The leakage rate past the plunger is 0.10 times the volume flowrate out of the needle.

2 leak

1

Q QV

A

Example 5.9 A bathtub is being filled with water from a faucet. The rate of flow from the faucet is steady at 9 gal/min. The tub volume is approximated by a rectangular space as indicated in Figure. Estimate the time rate of change of the depth of water in the tub, h/t, in in./min at any instant. Solve example using a deforming control volume that includes only the water accumulating in the buthtub.

water water2 2

Answer: 1.44 in./min10 ft 10 ftj

Q Qh

t A

Newton’s Second Law – Linear Momentum Equation

Time rate of change of the linear momentum of the system equals sum of external forces acting on the system

When control volume is coincident with a system at an instant of time:

Reynolds transport theorem with Bsys = system momentum and b = V, becomes

Time rate of change of the linear momentum of the system equals the time rate of change of the linear momentum of the contents of the control volume plus net rate of flow of linear momentum trough the control surface

Linear momentum equation is

Both surface and body forces act on the contents of the control volume

sys cv cs

DdV dV dA

Dt t

ˆV V V V n

cvcv csdV dA

t

ˆV V V n F

syssys

DdV

Dt V F

sys cv F F

Example 5.10 A horizontal jet of water exits a nozzle with a uniform speed of V1 = 10 ft/s,

strikes a vane, and is turned through an angle . Determine the anchoring force needed to hold the vane stationary. Neglect gravity and viscous effects.

21 1

21 1

o

Answer: 1 cos 11.64 1 cos lb

sin 11.64sin lb

=0 0; 0

90 11.64 lb; 11.64 lb

=

Ax

Az

Ax Az

Ax Az

F AV

F AV

F F

F F

o180 23.3 lb; 0 Ax AzF F

Example 5.11 Determine the anchoring force required to hold in place a conical nozzle attached to the end of a laboratory sink faucet when the water flowrate is 0.6 liter/s. The nozzle mass is 0.1 kg. The nozzle inlet and exit diameters are 16 mm and 5 mm, respectively. The nozzle axis is vertical and the axial distance between sections (1) and (2) is 30 mm. The pressure at section (1) is 464 kPa.

Example 5.11 Determine the anchoring force required to hold in place a conical nozzle attached to the end of a laboratory sink faucet when the water flowrate is 0.6 liter/s. The nozzle mass is 0.1 kg. The nozzle inlet and exit diameters are 16 mm and 5 mm, respectively. The nozzle axis is vertical and the axial distance between sections (1) and (2) is 30 mm. The pressure at section (1) is 464 kPa.

1 2 1 1 2 2

77 8 NA n wF m w w W W p A p A .

Example 5.12 Water flows through a horizontal, 180° pipe bend. The flow cross-sectional area is constant at a value of 0.1 ft2 through the bend. The flow velocity everywhere in the bend is axial and 50 ft/s/ The absolute pressures at the entrance and exit of the bend are 30 psia and 24 psia, respectively. Calculate the horizontal (x and y) components of the anchoring force required to hold the bend in place.

1 2 1 1 2 2

1324 lbAyF m v v p A p A

Example 5.13 Air flows steadily between two cross sections in a long, straight portion of 4-in. inside diameter pipe as indicated in Figure, where the uniformly distributed temperature and pressure at each cross section are given. If the average air velocity at section (2) is 1000 ft/s, we found in Example 5.2 that the average air velocity at section (1) must be 219 ft/s. Assuming uniform velocity distributions at sections (1) and (2), determine the frictional force exerted by the pipe wall on the air flow between sections (1) and (2).

1 1 2 2 1 1 2 2

2 1 2 2 1

793 lb

x

x

u m u m R p A p A

R A p p m u u

Example 5.14 If the flow of Example 5.4 is vertically upward, develop an expression for the fluid pressure drop that occurs between sections (1) and (2).

2

1 1 2 2 1 1 2 2

21

1 21 1

Solution:

3

zA

z

w m w w dA p A R W p A

w R Wp p

A A

Example 5.15 A static thrust stand as sketched in Figure is to be designed for testing a jet engine. The following conditions are known for a typical test: Intake air velocity = 200 m/s; exhaust gas velocity = 500 m/s; intake cross-sectional area = 1 m2; intake static pressure = -22.5 kPa = 78.5 kPa (abs); intake static temperature = 268 K; exhaust static pressure = 0 kPa = 101 kPa (abs). Estimate the nominal thrust for which to design.

1 1 2 2 1 1 th 2 2 atm 1 2

th

Solution:

=83 kN

u m u m p A F p A p A A

F

Linear Momentum Equation for a Moving CV

For a moving nonderofming control volume

For a constant control volume velocity, Vcv and steady flow:

cv cv cvcv csdV dA

t

ˆW V W V W n F

cvcsdA ˆW W n F

Example 5.17 A vane on wheels moves with constant velocity V0 when a stream of water

having a nozzle exit velocity of V1 is turned 45° by the vane as indicated in Figure (a). Note

that this is the same moving vane considered in Section 4.4.6 earlier. Determine the

magnitude and direction of the force, F, exerted by the stream of water on the vane surface. The speed of the water jet leaving the nozzle is 100 ft/s, and the vane is moving to the right with a constant speed of 20 ft/s.

2 2

1 o

21 8 lb; 53 lb;

=57.3 lb;

67 6

x z

x z

z

x

R R

R R R

R

R

-

.

tan .

Moment-of-Momentum Equation

Moment-of-momentum equation relates torques and angular momentum flow for the contents of a control volume (derivation)

cvcv cs

dV dAt

ˆr V r V V n r F

Application of the Moment-of-Momentum Equation

Example 5.18

shaft

shaft

(a) 3 34 N m

(b) 1 24 N m

(c) 797 rpm

T

T

.

.

Application of the Moment-of-Momentum Equation

Angular momentum equation is used to obtain torque and power for rotating machines

shaf in in in out out out

shaf in in in out out out

shafshaf in in out out

t

t

tt

T m r V m r V

W m U V m U V

Ww U V U V

m

Example 5.19

Energy Equation

Energy equation involves stored energy, heat transfer, and work

One-dimensional energy equation for steady-in-the-mean flow:

or in terms of enthalpy

Equation is valid for incompressible and compressible flows

2 2

net shaftin net in

out in2 2

p V p Vm u gz u gz Q W

2

net shaftcv cs in net in2

p Ve dV u gz dA Q W

t

ˆV n

2 2

net shaftin net in

out in2 2

V Vm h gz h gz Q W

Example 5.20

Example 5.21

Example 5.22

Energy Equation vs. Bernoulli Equation

Comparison of one-dimensional steady-flow energy equation per unit mass for incompressible flow

and Bernoulli equation (frictionless flow)

shows that for flow with friction the term

represent the loss of an available energy.

Thus, energy equation is often written in terms of loss as

out in netin

0u u q

2 2out out in in

out in out in netin2 2

p V p Vgz gz u u q

2 2out out in in

out in2 2

p V p Vgz gz

2 2out out in in

out in loss2 2

p V p Vgz gz

Example 5.23

Energy Equation

For incompressible flow through pumps, blowers, fans, and turbines energy equation can be expressed as

It is called the mechanical energy equation or extended Bernoulli equation

2 2out out in in

out in shaftin

loss2 2

p V p Vgz gz w

Example 5.24

Energy Equation

Energy equation written in terms of energy per unit weight involves heads

2 2out out in in

out in2 2 s L

p V p Vz z h h

g g

Example 5.25

END OF CHAPTER

Supplementary slides

back

Inflow across a typical portion of the control surface

in inin cs cs

B bV dA b dA ˆcos V n

Derivation of the Moment-of-Momentum Equation

particle

particle

Apply Newton's second law to a particle of fuid

Form the moment of each side of above equation with respect to origin of coordinate system

Note that

DV

Dt

DV

Dt

DV

Dt

V F

r V r F

r V

particle

and

Thus, since

0

by combining above equations we obtain

D VDV

Dt Dt

D

Dt

DV

Dt

VrV r

rV

V V

r V r F

Derivation of the Moment-of-Momentum Equation

syssys

syssys

Use sum of both sides to obtain

Reverse order of integration and differentiation to get

For a control volume that instantaneously coincident with the system

DV

Dt

DV

Dt

r V r F

r V r F

sys cv

sys cv cs

For the system and coincident nondeforming fixed control volume Reynolds transport theorem:

Moment-of-momentum equation for a fixed nondeforming co

DdV dV dA

Dt t

r F r F

ˆr V r V r V V n

cvcv cs

ntrol volume

dV dAt

ˆr V r V V n r F

back

• Continuum hypothesis:

– fluid is made up of fluid particles;

– each particle contains numerous molecules;

– infinitesimal particles of a fluid are tightly packed together

• Thus, motion of a fluid is described in terms of fluid particles rather than individual molecules.

• This motion can be described in terms of the velocity and acceleration of the fluid particles

• At a given instant of time, description of any fluid property may be given as a function of fluid location

• Representation of fluid parameters as function of spatial coordinates is termed a field representation of the flow

• Fluid parameters are functions of position ant time. For example, temperature in the room is completely specified by temperature field

Velocity Field

, , ,T T x y z t

Example 5.16 A sluice gate across a channel of width b is shown in the closed and open positions in Figure. Is the anchoring force required to hold the gate in place larger when the gate is closed or when it is open?

2 2cs axial

shaftcv axial

2 2 shaft

shaft shaft 2 2

shaft 2 2

da r V m

T

r V m T

W T r V m

W U V m

V W U

ˆr V V n

r V

Example 5.26